# Chemistry practical

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#3

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Can anyone help to me with the molar mass and questions 1-3 please

**Lucky10**)Can anyone help to me with the molar mass and questions 1-3 please

Molar mass is just the relative atomic mass of all the atoms in the molecule added together. So if a molecule has say 1 carbon and 4 hydrogens the molar mass is 1(12) + 4(1) = 16.

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(Original post by

Moles=mass/molar mass

Molar mass is just the relative atomic mass of all the atoms in the molecule added together. So if a molecule has say 1 carbon and 4 hydrogens the molar mass is 1(12) + 4(1) = 16.

**SGHD26716**)Moles=mass/molar mass

Molar mass is just the relative atomic mass of all the atoms in the molecule added together. So if a molecule has say 1 carbon and 4 hydrogens the molar mass is 1(12) + 4(1) = 16.

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#5

(Original post by

I don't understand how you do that for ones I have?

**Lucky10**)I don't understand how you do that for ones I have?

So molar mass is 3(12) + 8(1) + 1(16) = 60

So moles is 2.67/60 = 0.0445 moles burnt

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(Original post by

For question 2b, you need to find the moles of propanol burned. You know that 2.67g was burnt. Now all you need is the molar mass of propanol. Propanol has 3 carbons, 8 hydrogen s and 1 oxygen.

So molar mass is 3(12) + 8(1) + 1(16) = 60

So moles is 2.67/60 = 0.0445 moles burnt

**SGHD26716**)For question 2b, you need to find the moles of propanol burned. You know that 2.67g was burnt. Now all you need is the molar mass of propanol. Propanol has 3 carbons, 8 hydrogen s and 1 oxygen.

So molar mass is 3(12) + 8(1) + 1(16) = 60

So moles is 2.67/60 = 0.0445 moles burnt

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#7

(Original post by

How would you work out the molar mass of fuel

**Lucky10**)How would you work out the molar mass of fuel

Carbon has a relative atomic mass of 12

Hydrogen has a relative atomic mass of 1

Oxygen has a relative atomic mass of 16

Propanol has 3 carbons 8 hydrogens and 1 oxygen. You simply add them together the relative atomic masses.

Before doing these questions I think you should read your textbook and notes carefully.

Propanol is the fuel and so is ethanol

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(Original post by

In the periodic table you have the elements and each element has two numbers on it. One is the atomic number and one is the relative atomic mass. You are interested in the relative atomic mass.

Carbon has a relative atomic mass of 12

Hydrogen has a relative atomic mass of 1

Oxygen has a relative atomic mass of 16

Propanol has 3 carbons 8 hydrogens and 1 oxygen. You simply add them together the relative atomic masses.

Before doing these questions I think you should read your textbook and notes carefully.

Propanol is the fuel and so is ethanol

**SGHD26716**)In the periodic table you have the elements and each element has two numbers on it. One is the atomic number and one is the relative atomic mass. You are interested in the relative atomic mass.

Carbon has a relative atomic mass of 12

Hydrogen has a relative atomic mass of 1

Oxygen has a relative atomic mass of 16

Propanol has 3 carbons 8 hydrogens and 1 oxygen. You simply add them together the relative atomic masses.

Before doing these questions I think you should read your textbook and notes carefully.

Propanol is the fuel and so is ethanol

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#11

(Original post by

How would you do 2a and C

**Lucky10**)How would you do 2a and C

2c) q divided by moles.

Do you have a textbook? If you don't get this one, assuming you do OCR https://www.amazon.co.uk/gp/aw/d/019...nQL&ref=plSrch

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(Original post by

2a) q=mxcxdeltat

2c) q divided by moles.

Do you have a textbook? If you don't get this one, assuming you do OCR https://www.amazon.co.uk/gp/aw/d/019...nQL&ref=plSrch

**SGHD26716**)2a) q=mxcxdeltat

2c) q divided by moles.

Do you have a textbook? If you don't get this one, assuming you do OCR https://www.amazon.co.uk/gp/aw/d/019...nQL&ref=plSrch

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#13

Hi there,

The answer to question one is:

Propanol= C3H7OH+10/2 O2----->3CO2+ 4H20

Ethanol=C2H5OH+7/2O2----->2CO2+3H20

How I worked it out:

The general formula for an alcohol is the same formula for an alkane (cnh2n+2) But you're just replacing one of the hydrogen with an OH group (you will understand it better when you gt on to organic chemistry.) The products for combustion are always CO2 and H20 (Which you should learn in enthalpy change of combustion topic.) and the hydrocarbon is only reacting completely with oxygen. You can write 5O2 if it's easier for you I just prefer to balance with fractions.

The answer to question two is:

25080J

The equation you use for this is Q=MCT where M stands for the mass of your substance (in this case,water.), C stands for heat capacity and T is the temerature change.

Answer to question 2b is:

0.0445 (Hopefully, now that you know the formula for propanol you can work out why.)

2c: -563.60

The equation for this is enthalpy change= -q/n

Q is what you worked out in 2a.

(I'm also doing AS levels so this is good revision for me lol.)

The answer to question one is:

Propanol= C3H7OH+10/2 O2----->3CO2+ 4H20

Ethanol=C2H5OH+7/2O2----->2CO2+3H20

How I worked it out:

The general formula for an alcohol is the same formula for an alkane (cnh2n+2) But you're just replacing one of the hydrogen with an OH group (you will understand it better when you gt on to organic chemistry.) The products for combustion are always CO2 and H20 (Which you should learn in enthalpy change of combustion topic.) and the hydrocarbon is only reacting completely with oxygen. You can write 5O2 if it's easier for you I just prefer to balance with fractions.

The answer to question two is:

25080J

The equation you use for this is Q=MCT where M stands for the mass of your substance (in this case,water.), C stands for heat capacity and T is the temerature change.

Answer to question 2b is:

0.0445 (Hopefully, now that you know the formula for propanol you can work out why.)

2c: -563.60

The equation for this is enthalpy change= -q/n

Q is what you worked out in 2a.

(I'm also doing AS levels so this is good revision for me lol.)

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(Original post by

Hi there,

The answer to question one is:

Propanol= C3H7OH+10/2 O2----->3CO2+ 4H20

Ethanol=C2H5OH+7/2O2----->2CO2+3H20

How I worked it out:

The general formula for an alcohol is the same formula for an alkane (cnh2n+2) But you're just replacing one of the hydrogen with an OH group (you will understand it better when you gt on to organic chemistry.) The products for combustion are always CO2 and H20 (Which you should learn in enthalpy change of combustion topic.) and the hydrocarbon is only reacting completely with oxygen. You can write 5O2 if it's easier for you I just prefer to balance with fractions.

The answer to question two is:

25080J

The equation you use for this is Q=MCT where M stands for the mass of your substance (in this case,water.), C stands for heat capacity and T is the temerature change.

Answer to question 2b is:

0.0445 (Hopefully, now that you know the formula for propanol you can work out why.)

2c: -563.60

The equation for this is enthalpy change= -q/n

Q is what you worked out in 2a.

(I'm also doing AS levels so this is good revision for me lol.)

**noor.m**)Hi there,

The answer to question one is:

Propanol= C3H7OH+10/2 O2----->3CO2+ 4H20

Ethanol=C2H5OH+7/2O2----->2CO2+3H20

How I worked it out:

The general formula for an alcohol is the same formula for an alkane (cnh2n+2) But you're just replacing one of the hydrogen with an OH group (you will understand it better when you gt on to organic chemistry.) The products for combustion are always CO2 and H20 (Which you should learn in enthalpy change of combustion topic.) and the hydrocarbon is only reacting completely with oxygen. You can write 5O2 if it's easier for you I just prefer to balance with fractions.

The answer to question two is:

25080J

The equation you use for this is Q=MCT where M stands for the mass of your substance (in this case,water.), C stands for heat capacity and T is the temerature change.

Answer to question 2b is:

0.0445 (Hopefully, now that you know the formula for propanol you can work out why.)

2c: -563.60

The equation for this is enthalpy change= -q/n

Q is what you worked out in 2a.

(I'm also doing AS levels so this is good revision for me lol.)

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**noor.m**)

Hi there,

The answer to question one is:

Propanol= C3H7OH+10/2 O2----->3CO2+ 4H20

Ethanol=C2H5OH+7/2O2----->2CO2+3H20

How I worked it out:

The general formula for an alcohol is the same formula for an alkane (cnh2n+2) But you're just replacing one of the hydrogen with an OH group (you will understand it better when you gt on to organic chemistry.) The products for combustion are always CO2 and H20 (Which you should learn in enthalpy change of combustion topic.) and the hydrocarbon is only reacting completely with oxygen. You can write 5O2 if it's easier for you I just prefer to balance with fractions.

The answer to question two is:

25080J

The equation you use for this is Q=MCT where M stands for the mass of your substance (in this case,water.), C stands for heat capacity and T is the temerature change.

Answer to question 2b is:

0.0445 (Hopefully, now that you know the formula for propanol you can work out why.)

2c: -563.60

The equation for this is enthalpy change= -q/n

Q is what you worked out in 2a.

(I'm also doing AS levels so this is good revision for me lol.)

1

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**noor.m**)

Hi there,

The answer to question one is:

Propanol= C3H7OH+10/2 O2----->3CO2+ 4H20

Ethanol=C2H5OH+7/2O2----->2CO2+3H20

How I worked it out:

The general formula for an alcohol is the same formula for an alkane (cnh2n+2) But you're just replacing one of the hydrogen with an OH group (you will understand it better when you gt on to organic chemistry.) The products for combustion are always CO2 and H20 (Which you should learn in enthalpy change of combustion topic.) and the hydrocarbon is only reacting completely with oxygen. You can write 5O2 if it's easier for you I just prefer to balance with fractions.

The answer to question two is:

25080J

The equation you use for this is Q=MCT where M stands for the mass of your substance (in this case,water.), C stands for heat capacity and T is the temerature change.

Answer to question 2b is:

0.0445 (Hopefully, now that you know the formula for propanol you can work out why.)

2c: -563.60

The equation for this is enthalpy change= -q/n

Q is what you worked out in 2a.

(I'm also doing AS levels so this is good revision for me lol.)

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#17

(Original post by

And for propanol 2C3H7OH + 9 O2 >> 6 CO2 + 8H2O

**Lucky10**)And for propanol 2C3H7OH + 9 O2 >> 6 CO2 + 8H2O

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(Original post by

I think there's two ways of balancing it?

**noor.m**)I think there's two ways of balancing it?

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(Original post by

I think there's two ways of balancing it?

**noor.m**)I think there's two ways of balancing it?

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#20

(Original post by

Wb q4??

**Lucky10**)Wb q4??

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